CF1063B Labyrinth

思路：

使用bfs，不过每个点可以访问多次。

实现：

#include <iostream>
#include <queue>
#include <cstring>
using namespace std;

typedef pair<int, int> pii;

const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const int INF = 0x3f3f3f3f;

char a[2005][2005];
int n, m, r, c, L, R;
int d[2005][2005];

int main()
{
    while (cin >> n >> m >> r >> c >> L >> R)
    {
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                cin >> a[i][j];
                d[i][j] = INF;
            }
        }
        r--; c--;
        queue<pii> q;
        q.push(pii(r, c));
        d[r][c] = 0;
        while (!q.empty())
        {
            pii tmp = q.front();
            q.pop();
            int x = tmp.first, y = tmp.second;
            for (int i = 0; i < 4; i++)
            {
                int nx = x + dx[i], ny = y + dy[i];
                if (nx >= 0 && nx < n && ny >= 0 && ny < m && a[nx][ny] == '.')
                {
                    int nd = d[x][y] + (i == 3);
                    if (nd < d[nx][ny])
                    {
                        q.push(pii(nx, ny));
                        d[nx][ny] = d[x][y] + (i == 3);
                    }
                }
            }
        }
        int cnt = 0;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                if (d[i][j] <= L && d[i][j] + j - c <= R) cnt++;
            }
        }
        cout << cnt << endl;
    }
    return 0;
}